IB Analysis and Approaches HL · Integral calculus · Section 1 of 13

Antiderivatives and the indefinite integral

3 worked examples · By Álvaro López

Key idea

Integrating is differentiating in reverse: given a function, you look for another whose derivative is the first. Almost all the techniques in this topic (substitution, parts, partial fractions) end up reducing to one of the standard integrals in this section, so it is worth mastering them the way you know your times tables.

Antiderivative of a function

For example, $F (x) = x^{2}$ is an antiderivative of $f (x) = 2 x$ , but so are $x^{2} + 3$ and $x^{2} - \frac{1}{2}$ : when you differentiate, the constant disappears. Two antiderivatives of the same function can only differ by a constant, because the only function with derivative $0$ on an interval is a constant.

The indefinite integral and the constant $C$

Geometrically, the constant $C$ shifts the graph up or down: all the antiderivatives of $f$ are the same curve translated vertically.

Parabolas y = x² + C, all the antiderivatives of f(x) = 2x; the orange one, with C = -1, passes through (1, 0)

If the question gives a point that the antiderivative passes through (an initial condition), substitute that point into $F (x) + C$ and solve for $C$ . This picks out a single curve from the family.

Properties

Integration is linear, just like differentiation:

A number can be taken outside the integral, and you can integrate term by term. You cannot take out a function of $x$ or split a product: $\int f (x) g (x) d x \neq = \int f \cdot \int g$ .

Table of standard integrals

Each row comes from reading a row of the table of derivatives in topic 10 in reverse. The version with an inner function $f (x)$ is given: if its derivative $f^{'} (x)$ appears multiplying, the antiderivative is the usual one evaluated at $f (x)$ .

With $f (x) = x$ (and $f^{'} = 1$ ) the basic versions come out: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$ , $\int \frac{1}{x} d x = ln ∣ x ∣ + C$ , $\int e^{x} d x = e^{x} + C$ , $\int cos x d x = sin x + C$ , and so on. Notice that $x^{- 1}$ is the exception to the power rule: it gives a logarithm.

Integrals of the form $f (a x + b)$

The most common case is when the inner function is linear, $a x + b$ . Then $f^{'} = a$ is constant, and you only need to compensate for that factor by dividing by $a$ .

For example, $\int cos (5 x - 2) d x = \frac{1}{5} sin (5 x - 2) + C$ and $\int \frac{d x}{3 x + 1} = \frac{1}{3} ln ∣ 3 x + 1 ∣ + C$ .

When the denominator is $a^{2} + x^{2}$ or $a^{2} - x^{2}$ with $a \neq = 1$ , take out the $a$ to get $1 + (x / a)^{2}$ or $1 - (x / a)^{2}$ : $\int \frac{d x}{a ^{2} + x ^{2}} = \frac{1}{a} arctan \frac{x}{a} + C, \int \frac{d x}{a ^{2} - x ^{2}} = arcsin \frac{x}{a} + C (a > 0) .$

Worked examples

Example 1 Basic

Term-by-term antiderivatives

Calculate (with $x > 0$ where needed):

a) $\int (4 x^{3} - 6 x + 5) d x$ .

b) $\int (\frac{1}{x ^{2}} + 6 x) d x$ .

c) $\int (2 e^{x} - 4 cos x + \frac{1}{x}) d x$ .

Solution

Integrate each term with its formula and add a single constant at the end.

a) $\int (4 x^{3} - 6 x + 5) d x = x^{4} - 3 x^{2} + 5 x + C$ .

b) Write $\frac{1}{x ^{2}} = x^{- 2}$ and $x = x^{1/2}$ : $\int (x^{- 2} + 6 x^{1/2}) d x = \frac{x ^{- 1}}{- 1} + 6 \cdot \frac{x ^{3/2}}{3/2} + C = - \frac{1}{x} + 4 x^{3/2} + C .$

c) $\int (2 e^{x} - 4 cos x + \frac{1}{x}) d x = 2 e^{x} - 4 sin x + ln x + C$ (with $x > 0$ the absolute value is unnecessary).

Example 2 Medium

Integrals of $f (a x + b)$ and basic quotients

Calculate:

a) $\int e^{- 4 x + 2} d x$ .

b) $\int \frac{d x}{2 x + 5}$ , with $x > - \frac{5}{2}$ .

c) $\int \frac{d x}{9 + x ^{2}}$ .

d) $\int \frac{d x}{1 - 4 x ^{2}}$ , with $∣ x ∣ < \frac{1}{2}$ .

Solution

a) The exponential with a linear inner function of gradient $- 4$ : divide by $- 4$ . $\int e^{- 4 x + 2} d x = - \frac{1}{4} e^{- 4 x + 2} + C .$

b) A logarithm with an inner function of gradient $2$ : $\int \frac{d x}{2 x + 5} = \frac{1}{2} ln (2 x + 5) + C .$

c) Take out the $9$ : $\frac{1}{9 + x ^{2}} = \frac{1}{9} \cdot \frac{1}{1 + ( x /3 ) ^{2}}$ . Since $(\frac{x}{3})^{'} = \frac{1}{3}$ , you must compensate with another factor $3$ : $\int \frac{d x}{9 + x ^{2}} = \frac{1}{9} \cdot 3 arctan \frac{x}{3} + C = \frac{1}{3} arctan \frac{x}{3} + C .$

d) With $f (x) = 2 x$ and $f^{'} (x) = 2$ : $\frac{1}{1 - 4 x ^{2}} = \frac{1}{2} \cdot \frac{2}{1 - ( 2 x ) ^{2}}$ , so $\int \frac{d x}{1 - 4 x ^{2}} = \frac{1}{2} arcsin (2 x) + C .$

Example 3 Hard

Recovering a function from its derivative

The derivative of a function $f$ is $f^{'} (x) = \frac{4 x}{x ^{2} + 1} + 2 e^{- 2 x},$ and the graph of $f$ passes through the point $(0, 3)$ .

a) Find $f (x)$ .

b) Calculate $f (1)$ .

c) Find the equation of the tangent to the graph of $f$ at $x = 0$ .

Solution

a) The first term is of the form $\frac{g ^{'}}{g}$ with $g = x^{2} + 1$ ( $g^{'} = 2 x$ ), so $\frac{4 x}{x ^{2} + 1} = 2 \cdot \frac{2 x}{x ^{2} + 1}$ . The second is an exponential with inner function $- 2 x$ , which is integrated by dividing by $- 2$ : $f (x) = 2 ln (x^{2} + 1) + 2 \cdot \frac{e ^{- 2 x}}{- 2} + C = 2 ln (x^{2} + 1) - e^{- 2 x} + C .$ (As $x^{2} + 1 > 0$ the absolute value is not needed.) With $f (0) = 3$ : $2 ln 1 - e^{0} + C = 3 \Rightarrow - 1 + C = 3 \Rightarrow C = 4.$ So $f (x) = 2 ln (x^{2} + 1) - e^{- 2 x} + 4$ .

b) $f (1) = 2 ln 2 - e^{- 2} + 4$ (approximately $5.251$ ).

c) The tangent passes through $(0, f (0)) = (0, 3)$ and its gradient is $f^{'} (0) = \frac{0}{1} + 2 e^{0} = 2$ . It is $y = 2 x + 3$ .

In the exam: there is a mark reserved for finding $C$ using the given condition. If you integrate without writing the $+ C$ , you lose it, and the error carries over into the following parts. Write down $C$ as soon as you integrate.

Common mistakes

Summary

An antiderivative of $f$ is an $F$ with $F^{'} = f$ ; all antiderivatives differ by a constant: $\int f = F + C$ .
Integration is linear: take out constants and add term by term; a product cannot be split.
Memorise the table: powers (with $ln ∣ x ∣$ for $x^{- 1}$ ), $e^{x}$ , $a^{x}$ , sine, cosine, $\frac{1}{cos ^{2} x}$ , arcsine and arctangent.
With $f (a x + b)$ : antiderivative in $a x + b$ , then divide by $a$ ; check by differentiating.
An initial condition fixes $C$ and picks out a single curve from the family.

Keep studying

Maths classes in Aldaia and online

Want to work on it together?

Free first consultation: we look at your level and your exam, and build a plan. Álvaro replies on WhatsApp.

Message us on WhatsApp

Integral	Result
$\int f (x)^{n} f^{'} (x) d x$ ( $n \neq = - 1$ )	$\frac{f ( x ) ^{n + 1}}{n + 1} + C$
$\int \frac{f ^{'} ( x )}{f ( x )} d x$	$ln ∣ f (x)∣ + C$
$\int e^{f (x)} f^{'} (x) d x$	$e^{f (x)} + C$
$\int a^{f (x)} f^{'} (x) d x$	$\frac{a ^{f (x)}}{ln a} + C$
$\int sin f (x) f^{'} (x) d x$	$- cos f (x) + C$
$\int cos f (x) f^{'} (x) d x$	$sin f (x) + C$
$\int \frac{f ^{'} ( x )}{cos ^{2} f ( x )} d x$	$tan f (x) + C$
$\int \frac{f ^{'} ( x )}{1 - f ( x ) ^{2}} d x$	$arcsin f (x) + C$
$\int \frac{f ^{'} ( x )}{1 + f ( x ) ^{2}} d x$	$arctan f (x) + C$

Antiderivatives and the indefinite integral

Antiderivative of a function

The indefinite integral and the constant C

Properties

Table of standard integrals

Integrals of the form f(ax+b)

Worked examples

Common mistakes

Summary

Want to work on it together?

The indefinite integral and the constant $C$

Integrals of the form $f (a x + b)$